Maths Ext 2 Predictions/Thoughts (1 Viewer)

notme123

Well-Known Member
Joined
Apr 14, 2020
Messages
997
Gender
Male
HSC
2021
My criticism stands.

But I want to downplay it. It's not a great drama.

He went through quite a bit of hefty algebra and I'm glad he was the one doing it, not me.

Overall I think he did a great job.
how does one even get chosen to do this.
1. go to a prestigious school which will pay the smh to do this article
2. become ranked first in x2
3. profit
 

Siwel

Active Member
Joined
Aug 21, 2021
Messages
208
Gender
Male
HSC
2021
how does one even get chosen to do this.
1. go to a prestigious school which will pay the smh to do this article
2. become ranked first in x2
3. profit
on the facebook 2 days before they asked ppl, are you 1st in your school for x2?
 

5uckerberg

Well-Known Member
Joined
Oct 15, 2021
Messages
562
Gender
Male
HSC
2018
ah yes the classic article where boomers question why we still do things like this because it has no value to society. a pattern ive noticed is that top schools always get the opp to do this.
You took the words right out of my mouth.
 

s97127

Active Member
Joined
Mar 4, 2018
Messages
302
Gender
Male
HSC
2020
1637231783888.png
For this question, i want to prove 5^n -2^n - 3^n > 0 for n>= 2

(2+3)^n - 2^n - 3^n = 2^(n-1).3 + 2^(n-2).3^2 +....+ 3^(n-1) > 0 (QED)

Is this proof correct?
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,400
Gender
Male
HSC
2006
View attachment 33882
For this question, i want to prove 5^n -2^n - 3^n > 0 for n>= 2

(2+3)^n - 2^n - 3^n = 2^(n-1).3 + 2^(n-2).3^2 +....+ 3^(n-1) > 0 (QED)

Is this proof correct?
Almost...except you forgot the binomial coefficients.

Also, generally speaking, proving something is positive to prove it is non-zero doesn’t always work (it works in this case though). You can have expressions that can be positive or negative but is non-zero.
 

Eagle Mum

Well-Known Member
Joined
Nov 9, 2020
Messages
560
Gender
Female
HSC
N/A
Correct I got 100 for x1 but i got nowhere close to a state rank.
I imagine that by scaling extension students up, it’s quite compacted at the top, so amongst the cohort with a scaled mark of 100 would be a range of raw marks, but anyone who achieves a raw mark of 100 deserves to be equal first. Students in different schools who achieve a perfect score in the exam shouldn’t be ranked based on school assessments, since assessments aren’t standardised. Congratulations on your awesome achievement!
 

s97127

Active Member
Joined
Mar 4, 2018
Messages
302
Gender
Male
HSC
2020
Almost...except you forgot the binomial coefficients.

Also, generally speaking, proving something is positive to prove it is non-zero doesn’t always work (it works in this case though). You can have expressions that can be positive or negative but is non-zero.
Thanks. I forgot about coefficients. Regarding the other comment, it only works if all the terms are either greater/lesser than 0.
 

uart

Member
Joined
Dec 10, 2008
Messages
69
Gender
Male
HSC
N/A
same omission of detail as above in the locus question
Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative solution was invalid in Q14c(ii).


Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [] solution is also positive).

My reasoning is that where we subst,
,
in order to get the polynomial, we could equally well have substituted
,
to get the same equation.

So the polynomial is equally valid for solving for, , as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,

correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?
 
Last edited:

notme123

Well-Known Member
Joined
Apr 14, 2020
Messages
997
Gender
Male
HSC
2021
Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative solution was invalid in Q14c(ii).


Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [] solution is also positive).

My reasoning is that where we subst,
,
in order to get the polynomial, we could equally well have substituted
,
to get the same equation.

So the polynomial is equally valid for solving for, , as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,

correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?
i didnt justify it at all i think whoops
 

jyu

Member
Joined
Nov 14, 2005
Messages
623
Gender
Male
HSC
2006
Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative solution was invalid in Q14c(ii).


Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [] solution is also positive).

My reasoning is that where we subst,
,
in order to get the polynomial, we could equally well have substituted
,
to get the same equation.

So the polynomial is equally valid for solving for, , as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,

correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?
itute stated cos@=cospi/10=0.951approx so cannot be sqrt((5-sqrt5)/8)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top